# <u|A|u> ?

1. Jan 25, 2005

### trosten

How do u define this <u|A|u>? where A is a selfadjoint linear operator.

I have one book that defines is as ||A|u>||^2=<u|A|u>
(equation 2.6 http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.ps)

and since ||A|u>||^2 is the lenght of the vector A|u> squared that is equal to (<u|A)(A|u>).

I have another book (j.j sakurai) that defines it as (<u|)(A|u>)=(<u|A)(|u>)=<u|A|u> but this is the projection of A|u> on |u>.

It seems to me that the two definitions arent equal !? Any ideas?

2. Jan 25, 2005

### dextercioby

Of course,they're not equal,as they involve different "animals".

$$\langle \psi|\hat{A}|\psi\rangle$$

is the mean/avrerage of the observable A described by the selfadjoint operator $\hat{A}$ on the quantum state $|\psi \rangle$
,while the norm
$$|| \hat{A}|\psi\rangle ||^{2}$$

involves the matrix element of the SQUARE of the operator...

Daniel.

P.S.The operator is selfadjoint,that's why is the square...

Last edited: Jan 25, 2005
3. Jan 25, 2005

### trosten

I found the solution to the problem myself. If the operator is A=|n><n| then ofcourse AA = (|n><n|)(|n><n|) = |n><n|n><n| = |n><n| = A
Ofcourse repeated projections shouldnt change anything! silly me.

4. Jan 25, 2005

### dextercioby

OKay,u should have told right from the beginning that u're interested in true and IDEMPOTENT ortogonal projectors...

Daniel.

5. Jan 25, 2005

### trosten

haha yeah didnt think about that